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5u^2-3u-14=0
a = 5; b = -3; c = -14;
Δ = b2-4ac
Δ = -32-4·5·(-14)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{289}=17$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-17}{2*5}=\frac{-14}{10} =-1+2/5 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+17}{2*5}=\frac{20}{10} =2 $
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